A block of mass m is at rest on another block | vedclass

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(C) Let the total mass of the system be $M = 2m$. The equilibrium position is where the spring force balances the weight of both blocks: $k x_{eq} = 2

Vedclass · Accepted Answer

$\frac{2mg}{k}$

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(C) Let the total mass of the system be $M = 2m$. The equilibrium position is where the spring force balances the weight of both blocks: $k x_{eq} = 2mg$,so $x_{eq} = \frac{2mg}{k}$.For the blocks to remain in contact,the normal force $N$ between them must be greater than or equal to zero. For the upper block of mass $m$,the equation of motion is $mg - N = ma$,where $a$ is the downward acceleration. Thus,$N = m(g - a)$.For the blocks to remain in contact,we require $N \ge 0$,which implies $a \le g$. The maximum downward acceleration of the system in simple harmonic motion is $a_{max} = \omega^2 A$,where $\omega = \sqrt{\frac{k}{2m}}$.At the highest point of the oscillation,the acceleration is directed downwards with magnitude $\omega^2 A$. The condition $a \le g$ must hold at this point. However,the blocks will lose contact if the acceleration exceeds $g$. The limiting case is $a = g$,which occurs when the spring force is zero (at the natural length of the spring).At the equilibrium position,the spring is compressed by $x_{eq} = \frac{2mg}{k}$. If the amplitude $A$ is equal to $x_{eq}$,the block reaches the natural length of the spring at the highest point. Thus,the maximum amplitude is $A = \frac{2mg}{k}$.

$A$ block of mass $m$ is at rest on another block of the same mass as shown in the figure. The lower block is attached to a spring. What is the maximum amplitude of motion such that both blocks remain in contact?

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